Let → a and → b be two vectors such that | 2 → a + 3 → b | = | 3 →...

Given that

$\dfrac{1}{8}\overset{\to}{a}$ is a unit vector.

$| \overset{\to}{a} | = 8$
$| 2\overset{\to}{a} + 3\overset{\to}{b}{|}^{2} = | 3\overset{\to}{a} + \overset{\to}{b}{|}^{2}$

We know that$\overset{\to}{a} \cdot \overset{\to}{b} = \left|\overset{\to}{a}\right|\left|\overset{\to}{b}\right|cos\theta,$
$⇒ 4 | \overset{\to}{a}{|}^{2} + 9 | \overset{\to}{b}{|}^{2} + 12 | \overset{\to}{a} | | \overset{\to}{b} | \cos 60 ° = 9 | \overset{\to}{a}{|}^{2} + | \overset{\to}{b}{|}^{2} + 6 | \overset{\to}{a} | | \overset{\to}{b} | \cos 60 °$
$⇒ 5 | \overset{\to}{a}{|}^{2} = 8 | \overset{\to}{b}{|}^{2} + 6 | \overset{\to}{a} | | | \overset{\to}{b} ∣ \cos 60 °$
$⇒ 5\left(64\right) = 8 | \overset{\to}{b}{|}^{2} + 6\left(8\right) \cdot | \overset{\to}{b} | \dfrac{1}{2}$
\left⇒ | \overset{\to}{b}{|}^{2} + 3 | \overset{\to}{b} | - 40 = 0 ⇒ \left(\left|\overset{\to}{b}\right| + 8\right) | \overset{\to}{b} | - 5\right) = 0$<br />$⇒ | \overset{\to}{b} | = 5$